Ideal Gas on an Isothermal Cycle!

a) Use the ideal gas equation and solve for the variables required in each case.

b) Use the definition of work, to find the work. Use directly the equation for the change of internal energy to get it. And use the first law of thermodynamics to get the heat.

c) Sum all the works found before.

d) Use the equation of efficiency with the values found in b).

a) The ideal gas law states:

\begin{equation*}
PV = nRT,
\end{equation*}

State B: With the given values, solving for \(V_B\) we get:

\begin{equation*}
V_B = 0.022 \, \text{m}^3.
\end{equation*}

State A: The volume is the same as in A. Then:

\begin{equation*}
V_A = 0.022 \, \text{m}^3.
\end{equation*}

Solving for \(T_A\) from the ideal gas law we get:

\begin{equation*}
T_A = 133.3 \, \text{K}.
\end{equation*}

State C: By the ideal gas law, solving for the volume we get:

\begin{equation*}
V_C = 0.066 \, \ \text{m}^3.
\end{equation*}

Since from \(B\) to \(C\) the gas undergoes an isothermal process, then:

\begin{equation*}
T_C = 400 \, \text{K}.
\end{equation*}

b) The work definition is:

\begin{equation*}
W = \int_{V_i}^{V_f} P dV.
\end{equation*}

The change of internal energy is:

\begin{equation*}
\Delta U = \frac{5}{2} nR \Delta T.
\end{equation*}

The first law of thermodynamics states:

\begin{equation*}
\Delta U = Q – W.
\end{equation*}

Using those equation for each process we get:

Process A \(\rightarrow\) B.

\begin{equation*}
\Delta U_{A \rightarrow B} = 11086\, \text{J}.
\end{equation*}

\begin{equation*}
W_{A \rightarrow B} = 0.
\end{equation*}

\begin{equation*}
Q_{A \rightarrow B} = 11086\, \text{J}.
\end{equation*}

Process B \(\rightarrow\) C.

\begin{equation*}
\Delta U_{B \rightarrow C} = 0.
\end{equation*}

\begin{equation*}
W_{B \rightarrow C} = 7307.5 \, \text{J}.
\end{equation*}

\begin{equation*}
Q_{B \rightarrow C} = 7307.5 \, \text{J}.
\end{equation*}

Process C \(\rightarrow\) A.

\begin{equation*}
\Delta U_{C \rightarrow A} = -11086 \, \text{J}.
\end{equation*}

\begin{equation*}
W_{C \rightarrow A} = -4434.4 \, \text{J}.
\end{equation*}

\begin{equation*}
Q_{C \rightarrow A} = – 15520.4 \, \text{J}.
\end{equation*}

c) By the sum of the three works found in b) we get:

\begin{equation*}
W_{ \text{Total}} = 2873.1 \, \text{J}.
\end{equation*}

d) The efficiency can be written as:

\begin{equation*}
\epsilon = \frac{W_{ \text{Total}}}{Q_{\text{abs}}}.
\end{equation*}

The total work was already found. The heat absorbed is \(Q_{A \rightarrow B} + Q_{B \rightarrow C}\). Then:

\begin{equation*}
\epsilon = 0.156.
\end{equation*}

For a more detailed explanation of any of these steps, click on “Detailed Solution”.

Let \(n = 2\) be the number of moles of gas. Also, let \(P_0 = 10^{5}\) Pa be the pressure at states A and C.

a) Some of the state variables (pressure, volume, and temperature) are already given in the figure in the prompt. We can find the rest by using the ideal gas law.

The ideal gas law states that for \(n\) moles of an ideal gas, its pressure \(P\), volume \(V\), and temperature \(T\) are related by the equation:

\begin{equation}
\label{EQ:GAS}
PV = nRT,
\end{equation}

where \(R\) is the ideal gas constant. Let’s consider each state separately:
State B: We are given the temperature \(T_B = 400 \ K\), and the pressure \(P_B = 3 P_0 = 3\cdot 10^5\ \text{Pa}\). If we divide eq. \eqref{EQ:GAS} by \(P\), we find that the volume for an ideal gas is given by

\begin{equation}
\label{EQ:V}
V = \frac{nRT}{P},
\end{equation}

and substituting \(T_B\) and \(P_B\) gives

\begin{equation}
\label{EQ:VB}
V_B = \frac{nRT_B}{P_B}
= \frac{nRT_B}{3P_0}
= \left(2\ \text{mol}\right) \left(8.3145 \frac{\ \text{J}}{\ \text{mol}\ \ \text{K}}\right)\frac{400\ \ \text{K}}{3\times10^5\ \ \text{Pa}}
= 0.022\ \ \text{m}^3.
\end{equation}

State C: Since the gas undergoes an isothermal process between states B and C, we have

\begin{equation}
\label{EQ:TC}
T_C = T_B = 400\ K.
\end{equation}

Also, from the figure, we know that \(P_C = P_0 = 10^5\ \ \text{Pa}\). Hence, from eq. \eqref{EQ:V}, the volume \(V_C\) is given by

\begin{equation}
\label{EQ:VC}
V_C = \frac{nRT_C}{P_C}
= \frac{nRT_B}{P_0}
= 0.066\ \ \text{m}^3.
\end{equation}

State A: From the figure, we know that \(P_A = P_0 = 10^5\ \ \text{Pa}\). Also, we can notice that \(V_A = V_B\). Hence, from eq. \eqref{EQ:VB}, we get

\begin{equation}
\label{EQ:VA}
V_A = \frac{nRT_B}{3P_0}
= 0.022\ \ \text{m}^3,
\end{equation}

and we can now find the temperature \(T_A\) using the ideal gas law. If we divide eq. \eqref{EQ:GAS} by \(nR\), we obtain that the temperature for an ideal gas is given by

\begin{equation}
\label{EQ:T}
T = \frac{PV}{nR}.
\end{equation}

Substituting eq. \eqref{EQ:VA}, and \(P_A = P_0\) gives

\begin{equation}
\label{EQ:TA}
T_A = P_0 \left(\frac{nRT_B}{3P_0}\right)\left(\frac{1}{nR}\right)
= \frac{T_B}{3}
= \frac{400\ \ \text{K}}{3} = 133.3\ \ \text{K}.
\end{equation}
(b) We can find the internal energy of each state directly from the temperatures found in part (a) by using the equipartition theorem. We can then use these internal energies to find the change in internal energy during each process. Next, we can use the area under the curve of the P-V diagram to find the work done by the gas during each process. Finally, we can use the first law of thermodynamics to find the heat exchanged during each process.

Internal Energy

The equipartition theorem states that for \(n\) moles of a gas in equilibrium, there is an average internal energy of \(\frac{1}{2} nRT\) for each degree of freedom. A diatomic gas has 5 degrees of freedom because for each molecule, there is translational Kinetic energy in three dimensions, and rotational Kinetic energy over two rotation axes. Therefore, the internal energy \(U\) of \(n\) moles of a diatomic gas is given by

\begin{equation}
\label{EQ:U}
U = 5 \left(\frac{1}{2} nRT\right) = \frac{5}{2} nRT.
\end{equation}

We can substitute the temperatures found in part (a) in this equation to obtain expressions for the internal energy at all four states. Let’s consider each process separately:

Process A \(\rightarrow\) B: The change in internal energy is defined as

\begin{equation}
\label{EQ:pre_DUAB}
\Delta U_{A \rightarrow B} = U_B – U_A.
\end{equation}

Now, according to eq. \eqref{EQ:U}, the internal energies \(U_B\) and \(U_A\) are given by

\begin{equation}
U_A = \frac{5}{2} nRT_A,
\end{equation}

and

\begin{equation}
U_B = \frac{5}{2} nRT_B.
\end{equation}

Substituting eq. \eqref{EQ:TA} yields

\begin{equation}
\label{EQ:UA}
U_A = \frac{5}{2} nR \frac{T_B}{3}
= \left(\frac{5}{6}\right) nR T_B.
\end{equation}

and

\begin{equation}
\label{EQ:UB}
U_B = \frac{5}{2} nR T_B.
\end{equation}

Therefore, inserting these equations in eq. \eqref{EQ:pre_DUAB} gives

\begin{equation}
\label{EQ:DUAB}
\Delta U_{A \rightarrow B} = \frac{5}{2} nR T_B – \frac{5}{6} nR T_B = \frac{5}{2} nRT_B\left( 1 – \frac{1}{3}\right) = \frac{5}{2}nRT_B\left(\frac{2}{3}\right)
= \frac{5}{3} nRT_B,
\end{equation}

and inserting numerical values yields

\begin{equation}
\Delta U_{A \rightarrow B} = \frac{5}{3} \left(2\ \text{mol}\right) \left(8.3145 \ \frac{J}{\text{mol}\ \text{K}}\right)\left(400\ \text{K}\right)
= 11086 \, \text{J}.
\end{equation}
Process B \(\rightarrow\) C: From eq. \eqref{EQ:U}, the internal energy is determined by the temperature of the gas. Since the process \(B \rightarrow C\) is isothermal, we have \(T_C = T_B\), and thus we have \(U_B = U_C\). Therefore, from eq. \eqref{EQ:UB}, we have

\begin{equation}
\label{EQ:UC}
U_C = \frac{5}{2} nRT_B,
\end{equation}

and thus,

\begin{equation}
\label{EQ:DUBC}
\Delta U_{B\rightarrow C} = U_C – U_B = 0.
\end{equation}

Process C \(\rightarrow\) A: From eqs. \eqref{EQ:UA} and \eqref{EQ:UC}, we get

\begin{equation}
\label{EQ:DUCA}
\Delta U_{C\rightarrow A} = U_A – U_C = \frac{5}{6} nR T_B – \frac{5}{2} nRT_B = \frac{5}{2} nRT_B \left(\frac{1}{3} – 1\right) = \frac{5}{2} nRT_B \left(-\frac{2}{3}\right)
\end{equation}

and substituting numerical values gives
\begin{equation}
\Delta U_{C\rightarrow A} = – \frac{5}{3} \left(2\ \ \text{mol}\right) \left(8.3145 \ \frac{\text{J}}{\ \text{mol}\ \text{K}}\right)\left(400\ \text{K}\right)
= – 11086 \, \text{J}.
\end{equation}

Work

The work performed by the gas \(W\) during a process can be shown to be equal to the area under the curve of the line describing the process in a P-V diagram. This can be formally written as

\begin{equation}
\label{EQ:W}
W = \int_{V_i}^{V_f} P dV.
\end{equation}

Let’s consider each process separately:

Process A \(\rightarrow\) B: The area under the curve described by this process is zero (i.e. the volume does not change during this process), as shown in figure 1.

Hence,

\begin{equation}
\label{EQ:WAB}
W_{A \rightarrow B} = 0.
\end{equation}

These results can be also obtained from eq. \eqref{EQ:W} because both limits of integration are equal, and so the integral is zero.

Process B \(\rightarrow\) C: In this case, the pressure is not constant. Hence, we must perform the integral in eq. \eqref{EQ:W} in order to find the work done by the gas. First, we must write the pressure as a function of the volume. This can be done by using the ideal gas law. If we divide eq. \eqref{EQ:GAS} by \(V\), we get that for any ideal gas

\begin{equation}
P = \frac{nRT}{V}.
\end{equation}

Now, during this particular process, the temperature remains constant at a value of \(T = T_B = 400 \ \ \text{K}\). Hence,

\begin{equation}
P = \frac{nRT_B}{V}
\end{equation}

and substituting this in eq. \eqref{EQ:W} gives

\begin{equation}
W_{B \rightarrow C} = \int_{V_i}^{V_f} \frac{nRT_B}{V} dV = nRT_B \int_{V_i}^{V_f} \frac{1}{V} dV,
\end{equation}

Solving the integral we get

\begin{equation}
W_{B \rightarrow C} = nRT_B V\Big|_{V_i}^{V_f} = nRT_B (\ln (V_f) – \ln(V_i)) = nRT_B \ln\left(\frac{V_f}{V_i}\right),
\end{equation}

where \(V_f = V_C\) and \(V_i = V_B\). Substituting these volumes from eqs. \eqref{EQ:VB} and \eqref{EQ:VC} gives

\begin{equation}
W_{B \rightarrow C} = nRT_B \ln\left(\frac{V_C}{V_B}\right) = nRT_B \ln\left( \frac{\left(\frac{nRT_B}{P_0}\right) }{ \left(\frac{nRT_B}{3P_0}\right) }\right) = nRT_B \ln\left( \frac{1}{\frac{1}{3}}\right),
\end{equation}

which is the same as

\begin{equation}
\label{EQ:WBC}
W_{B \rightarrow C} = nRT_B \ln\left(3\right).
\end{equation}

Inserting numerical values, we have

\begin{equation}
W_{B \rightarrow C} = \left(2\ \text{mol}\right) \left(8.3145 \frac{\text{J}}{\text{mol}\ \text{K}}\right)\left(400\ \text{K}\right) \ln\left(3\right)
= 7307.5 \, \text{J}.
\end{equation}

Figure 2 illustrates the work for this part of the process:

Process C \(\rightarrow\) A: During this process, the volume changes from \(V_i = V_C\) to \(V_f = V_A\) at a constant pressure of \(P_0.\) Hence, from eq. \eqref{EQ:W}, we get

\begin{equation}
W_{C \rightarrow A} = \int_{V_C}^{V_A} P_0 dV = P_0 \int_{V_C}^{V_A} dV = P_0 V\Big|_{V_C}^{V_A} = P_0 (V_A – V_C),
\end{equation}

and substituting eqs. \eqref{EQ:VC} and \eqref{EQ:VA} yields

\begin{equation}
\label{EQ:WCA}
W_{C \rightarrow A} = P_0 \left(\frac{nRT_B}{3P_0}- \frac{nRT_B}{P_0} \right) = P_0 \frac{nRT_B}{P_0} \left(\frac{1}{3} – 1\right)
= -\frac{2}{3} nRT_B.
\end{equation}

Finally, after inserting numerical values, we obtain
\begin{equation}
W_{C \rightarrow A} = -\frac{2}{3} \left(2\ \text{mol}\right) \left(8.3145 \frac{\text{J}}{\text{mol}\ \text{K}}\right)\left(400\ \text{K}\right)
= -4434.4 \, \text{J}.
\end{equation}

We could have easily found this work graphically, as the area of the rectangle as shown in figure 3.

Heat

Now, we should find the heat exchanged during the three processes. Since we have already found the work and the change in internal energy of all processes, we can easily find the heat exchanged by using the first law of thermodynamics. This law states that for a substance undergoing a thermodynamic process, the change in internal energy equals the heat \(Q\) absorbed by the substance minus the work \(W\) done by it. This can be written as

\begin{equation}
\Delta U = Q – W,
\end{equation}

where a positive sign for \(Q\) indicates that the substance absorbed heat, and a negative sign indicates that it radiated heat. Similarly, when \(W > 0\), the substance did work during the process, and when \(W < 0\), work was done on the substance. Solving for \(Q\) yields \begin{equation} Q = \Delta U + W. \end{equation}

Process A \(\rightarrow\) B: From eqs. \eqref{EQ:DUAB} and \eqref{EQ:WAB} we get \begin{equation} Q_{A\rightarrow B} = \frac{5}{3} nRT_B + 0 = \frac{5}{3} nRT_B , \end{equation}

and putting numerical values

\begin{equation} \label{EQ:QAB} Q_{A\rightarrow B} = \frac{5}{3} \left(2\ \text{mol}\right) \left(8.3145 \frac{\text{J}}{\text{mol}\ \text{K}}\right)\left(400\ \text{K}\right) = 11086 \ \text{J}. \end{equation}

Process B \(\rightarrow\) C: From eqs. \eqref{EQ:DUBC} and \eqref{EQ:WBC}, we get \begin{equation} Q_{B\rightarrow C} = 0 + nRT_B \ln\left(3\right) = nRT_B \ln\left(3\right), \end{equation}

namely,

\begin{equation} \label{EQ:QBC} Q_{B\rightarrow C} = \left(2\ \text{mol}\right) \left(8.3145 \frac{\text{J}}{\text{mol}\ \text{K}}\right)\left(400\ \text{K}\right) \ln\left(3\right) = 7307.5 \, \text{J}. \end{equation}

Process C \(\rightarrow\) A: From eqs. \eqref{EQ:DUCA} and \eqref{EQ:WCA}, we get \begin{equation} Q_{C\rightarrow A} = – \frac{5}{3} nRT_B + \left(-\frac{2}{3} nRT_B\right) = -\frac{7}{3} nRT_B , \end{equation}

which is the same as

\begin{equation} \label{EQ:QCA} Q_{C\rightarrow A} = -\frac{7}{3} \left(2\ \text{mol}\right) \left(8.3145 \frac{\text{J}}{\text{mol}\ \text{K}}\right)\left(400\ \text{K}\right) = -15520.4 \, \text{J}. \end{equation}

Here, the negative sign indicates that the gas transferred heat into its surroundings.

(c) The total work \(W\) performed by the gas during a cycle is simply the sum of the works performed by the gas during each process. This can be written as \begin{equation} W = W_{A\rightarrow B} + W_{B\rightarrow C} + W_{C\rightarrow A}, \end{equation} and substituting eqs. \eqref{EQ:WAB}, \eqref{EQ:WBC}, and \eqref{EQ:WCA} gives \begin{equation} W = 0 + nRT_B \ln\left(3\right) + \left(-\frac{2}{3} nRT_B\right) = nRT_B \left(\ln\left(3\right) – \frac{2}{3}\right), \end{equation}

which is equivalent to

\begin{equation} \label{EQ:WTOTAL} W = \left(2\ \text{mol}\right) \left(8.3145 \frac{\text{J}}{\text{mol}\ \text{K}}\right)\left(400\ \text{K}\right) \left(\ln\left(3\right) – \frac{2}{3}\right) = 2873.1 \, \text{J}. \end{equation}

(d) The efficiency \(\epsilon\) of an engine performing a thermodynamic cycle is defined as the net work performed \(W\) divided by the heat absorbed by the gas \(Q_{\text{abs}}\). This can be written as: \begin{equation} \label{EQ:EFF} \epsilon = \frac{W}{Q_{\text{abs}}}. \end{equation} In this case, \(W\) is given by \eqref{EQ:WTOTAL}. In order to find \(Q_{\text{abs}}\), we should identify in which processes the gas absorbs heat and then sum up the amount of heat absorbed on each of these processes. From eqs. \eqref{EQ:QAB}, \eqref{EQ:QBC}, and \eqref{EQ:QCA}, we can notice that \(Q_{A\rightarrow B}, Q_{B\rightarrow C} > 0\), and \(Q_{C \rightarrow A} < 0\). As explained in part (b), we define our sign conventions so that \(Q>0\) indicates that the gas absorbed heat. Therefore,

\begin{equation}
Q_{\text{abs}} = Q_{A\rightarrow B} + Q_{B\rightarrow C},
\end{equation}

and substituting eqs. \eqref{EQ:QAB} and \eqref{EQ:QBC} gives

\begin{equation}
Q_{\text{abs}} = \frac{5}{3} nRT_B + nRT_B \ln\left(3\right)
= nRT_B \left(\frac{5}{3} + \ln\left(3\right)\right) .
\end{equation}

Hence, substituting this equation, and eq. \eqref{EQ:WTOTAL} in eq. \eqref{EQ:EFF} yields

\begin{equation}
\epsilon = \frac{W}{Q_{\text{abs}}} = nRT_B \frac{\ln(3) – \frac{2}{3}}{nRT_B (\frac{5}{3} + \ln(3)) }
= \frac{\ln(3) – \frac{2}{3}}{\frac{5}{3} + \ln(3)}
= 0.156 = 15.6 \%
\end{equation}

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